Question

17. Factorise : (x2 – 3x) (x² – 3x - 1) - 20.​

Answer 1

Answer:

(x²-3x+4)(x²-3x-5)

Step-by-step explanation:

Given expression is

(x²-3x)(x²-3x-1)-20

Let (x²-3x)= a.

So the expression becomes:

a(a-1)-20

= a²-a-20

= a²-5a+4a-20

= a(a-5)+4(a-5)

= (a-5)(a+4)

= (x²-3x-5)(x²-3x+4) [putting value of a=x²-3x]

= (x²-3x+4)(x²-3x-5).

So, required factorisation is (x²-3x+4)(x²-3x-5).(Ans)

Answer 2

Answer:

Question :-

➠ (x² - 3x) (x² - 3x - 1) - 20

Answer :-

➠ (x² - 3x) (x² - 3x - 1) - 20

➻ (x² - 3x) {(x² - 3x) - 1} - 20

➻ (p) {(p) - 1} - 20 [ Taking x² - 3x = p]

➻ p² - p - 20

➻ p² - (5 - 4)p - 20

➻ p² - 5p + 4p - 20

➻ p(p - 5) + 4(p - 5)

➻ (p - 5) (p + 4)

➻ (x² - 3x - 5) (x² - 3x + 4). [ Putting value of p = x² - 3x]

∴ The value of (x² - 3x) (x² - 3x - 1) - 20 is (x² - 3x - 5) (x² - 3x + 4)